I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.

Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} < 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$ where $\frac{p}{q}$ is the mentioned representative.

Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.

Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?

From the geometrical form ($e^{i\varphi}$ corresponds to angle $\varphi$) you have: $e^{(2\pi p)/q}=1$ implies $2\pi p/q=2\pi k$ for some $k\in\mathbb Z$; i.e. $\frac pq\in\mathbb Z$. Was this what you had problem with?
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Martin SleziakAug 4 '12 at 10:49

BTW you can read here how to reply in comments. (It was only accident that I came back to this question and I saw your comment directed to me. If you want to get attention of other users who previously left comment, you can use @username.)
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Martin SleziakAug 4 '12 at 13:23

This is what I thought. But somehow I wasn't sure that 1. Every q-th root of unity is of the form $e^{\frac{2 \pi i p}{q}}$ and consequently that for every n-th root of unity z, $e^k=1$ if and only if $n|k$. As for the well defined remark isn't that implied by the uniqueness of the representative?
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JernejAug 4 '12 at 18:40

You have a morphism of groups $ex:\mathbb R \to S^1: r\mapsto e^{2i\pi r} $, where $S^1$ is the multiplicative group of complex numbers with $\mid z\mid=1$. This morphism is surjective and has kernel $\mathbb Z$.
[The wish to have kernel $\mathbb Z$ instead of $ 2\pi \mathbb Z$ dictated the choice of $ex(r)=e^{2i\pi r}$ instead of $e^{ir}$].

Restricting the morphism to $\mathbb Q$ induces a morphism $res(ex):\mathbb Q\to S^1$ with kernel $\mathbb Q\cap \mathbb Z=\mathbb Z$ and image $\mu_\infty\stackrel {def}{=} e^{2i\pi \mathbb Q} \subset S^1$.
The crucial observation is that this image is $\mu_\infty=\bigcup_n \mu_n$, where $\mu_n$ is the set of $n$-roots of unity $e^{\frac {2i\pi k}{n}}\quad (k=1,2,...,n)$.
Hence $\mu_\infty$ is the set of all roots of unity i.e. the set of complex numbers $z \in \mathbb C$ with $z^n=1$ for some $n\in \mathbb N^*.$

Applying Noether's isomorphism you finally get the required group isomorphism (be attentive to the successive presence and absence of a bar over the $q$ in the formula)
$$Ex: \mathbb Q/\mathbb Z \xrightarrow {\cong} \mu_\infty:\overline {q}\mapsto e^{2i\pi q} $$